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If the initial concentration of AB is 0.120 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 83s?

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Final Answer:

After 83 seconds, the concentration of A is 0.0729 M and the concentration of B is 0.0729 M.

Step-by-step explanation:

The reaction is:

AB(aq) → A(aq) + B(aq)

The rate law for this reaction is:

rate = k[AB]

where:

k is the rate constant

[AB] is the concentration of AB

We can use the following equation to relate the concentration of AB to the concentrations of A and B:

[AB] = [A] - [B]

where:

[A] is the concentration of A

[B] is the concentration of B

Substituting this equation into the rate law, we get:

rate = k([A] - [B])

We can also use the following equation to relate the change in the concentration of A to the change in the concentration of B:

Δ[A] = -Δ[B]

where:

Δ[A] is the change in the concentration of A

Δ[B] is the change in the concentration of B

Substituting this equation into the rate law, we get:

rate = k[A]

This equation is separable, which means that we can separate the variables and integrate both sides of the equation. The integrated form of the equation is:

ln[A] = -kt + ln[A0]

where:

[A] is the concentration of A at time t

[A0] is the initial concentration of A

k is the rate constant

t is the time

We can solve for k by substituting the following values into the equation:

[A] = 0.120 M

[A0] = 0.0729 M

t = 83 s

Solving for k, we get:

k = -0.00120 s^-1

Now we can use the equation to calculate the concentration of A after 83 seconds:

ln[A] = -(-0.00120 s^-1) * 83 s + ln(0.120 M)

Solving for [A], we get:

[A] = 0.0729 M

Therefore, after 83 seconds, the concentration of A is 0.0729 M and the concentration of B is 0.0729 M.

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